Understanding Real Analysis in Mathematics: The Lebesgue Integral and Measure Theory

Reasons for developing the Lebesgue integral and measure theory

We dive into "Real Analysis," an essential math course, with the help of measure theory to show how important the Lebesgue integral is for solving old math problems from before the 20th century. Especially, the mathematical theories in the Dirichlet principle in PDE and the Fourier representation deeply rely on the measure theory.

Our exploration begins by elucidating the advantages of the Lebesgue integral over its Riemann counterpart, using a simple example of Poisson's equation in three-dimensional space: $$- \nabla^2 u(\mathbf{r}) = \rho(\mathbf{r}),\quad \mathbf{r}=(x,y,z)\in \mathbb{R}^3,$$ subject to the boundary condition that $ u(\mathbf{r}) $ vanishes as $ \mathbf{r} $ approaches infinity.

To solve this problem intuitively, we utilize the Dirac delta function, symbolized by $\delta(\mathbf{r})$, to express the solution in a convolution format: $$u(\mathbf{r}) = \iiint \underbrace{\nabla^2 \Phi(\mathbf{r} - \mathbf{r'})}_{\delta(\mathbf{r} - \mathbf{r'})} \, u(\mathbf{r'}) \, d\mathbf{r'} = \iiint \Phi(\mathbf{r} - \mathbf{r'}) \underbrace{ \nabla^2 u(\mathbf{r'})}_{\rho(\mathbf{r'})} d\mathbf{r'},$$ where $ \Phi(\mathbf{r}) = \frac{1}{4\pi|\mathbf{r}|} $ represents the fundamental solution to the Laplace operator. Unfortunately, the middle integral above doesn't make sense in terms of the Riemann integral as the Dirac delta function, while not  the traditional definition of a function, is best understood within the framework of limits. Upon transitioning to the Lebesgue integral, we are also confronted with a notable incongruity:$$u(\mathbf{r}) = \nabla^2 \iiint \Phi(\mathbf{r} - \mathbf{r'}) \, u(\mathbf{r'}) \, d\mathbf{r'}\neq  \iiint \underbrace{ \nabla^2 \Phi(\mathbf{r} - \mathbf{r'})}_{0\mbox{  if  } \mathbf{r'} \neq \mathbf{r}}  u(\mathbf{r'}) \, d\mathbf{r'} =0.$$ Careful analysis is essential to grasp the meaning of the identity  $\nabla^2 \Phi(\mathbf{r} - \mathbf{r'})= \delta(\mathbf{r} - \mathbf{r'})$.  It's crucial to understand that the Laplacian, positioned outside the integral, cannot directly enter inside the integral.  In other ward, $$ \lim_{n\to \infty}\iiint \nabla^2\Phi_n(\mathbf{r} - \mathbf{r'}) \, u(\mathbf{r'}) \, d\mathbf{r'}\neq \iiint \lim_{n\to \infty}\nabla^2\Phi_n(\mathbf{r} - \mathbf{r'}) \, u(\mathbf{r'}) \, d\mathbf{r'}$$ where $ \Phi_n(\mathbf{r}) = \frac{1}{4\pi|\mathbf{r}|+1/n}.$ 

This complicated concept is elucidated in Real Analysis courses through the Lebesgue Dominated Convergence Theorem and measure theory.

Understanding Measures and Measurable Sets with ease

To grasp the intuitive concept of measure and measurability, let's focus on a set \( E \) in the two-dimensional space \( \mathbb{R}^2 \). Here, the area of set \( E \) will be denoted by \( \mu(E) \).
We can quantify the area \( \mu(E) \) by using rectangles:
  • If \( E = (a,b) \times (c,d) \) is a rectangle, then \( \mu(E) = (b-a)(d-c) \), making it measurable.
  •  For a sequence of rectangles \( E_j \) with \( E_j \cap E_k = \emptyset \) for \( j \neq k \), we have \( \mu\left(\bigcup_{j=1}^\infty E_j\right) = \sum_{j=1}^\infty \mu(E_j) \), showing that the union of rectangles is also measurable.
  •  It appears that any open domain \( E \) is measurable since an open set can typically be represented as a countable union of rectangles. Here, we must acknowledge the acceptability of countable sums in measurability.
  •  Consider rectangles \( E_n = (a,a+1/n) \times (c,c+1/n) \), where \( \mu(E_n) = 1/n^2 \). As \( \lim_{n\to \infty} \mu(E_n) = 0 \), we conclude that the measure of any single point is zero.
  • Based on the previous observation, it follows that the measure of any countable set of points also equals zero.
Measure theory provides a systematic framework for assigning a positive quantity, denoted by $\mu(E)$, to each suitable subset $E$, representing its area or measure. A triple $(X, \mathcal{M}, \mu)$ is defined as a measure space if it satisfies the following conditions:
  1. $X$ is a set.
  2. $\mathcal{M}$ is a $\sigma$-algebra over the set $X$, meaning it is closed under countable unions and complementations.
  3. $\mu$ is a measure on $\mathcal{M}$, adhering to the following properties:
    1. $\mu(E) \geq 0$ for any $E \in \mathcal{M}$.
    2. For all countable collections ${ E_j : j = 1, 2, \ldots }$ of pairwise disjoint sets in $\mathcal{M}$, $\mu\left(\bigcup_j E_j \right) = \sum_j \mu(E_j)$.
    3.  $\mu(\emptyset) = 0$.

Lebesgue Measurable Functions

To simplify our explanation, we limit our focus to the standard Lebesgue measure, denoted as $\mu$, within the space $X=\mathbb{R}^2$.

A function $f:X \to \mathbb{R}$ is measurable if $f^{-1}(U)$ is measurable for every open set $U$. Then, we have
  •  $f:X\to \mathbb{R}$ is measurable if $\{ x\in X~:~ f(x)>a\}\in\mathcal{M}$ for all $a\in\mathbb{R}$.
  •  If $f$ and $g$ are measurable, then so are $f+g$, $fg$, $\max\{ f, g\}$, $\min\{f,g\}$, and $|f|$.
  •   If $\{ f_j\}$ is a sequence of measurable functions, then $\lim\sup_j f_j$ and $\lim\inf_j f_j$ are measurable.
Let $\mathcal{M}_{\text{able}}$ denote the set of measurable functions. Let $\mathcal{S}_{\text{imple}}$ denote the set of all simple functions, which are finite linear combinations of characteristic functions: $\mathcal{S}_{\text{imple}}=\{\phi=\sum_{j=1}^n c_j\chi_{E_j} : E_j\in\mathcal{M},\,c_j\in\mathbb{R}\}.$ Here, $\chi_E$ represents the characteristic function of set $E$. For any function $f\in \mathcal{M}_{\text{able}}$, we say $f=0$ holds almost everywhere (abbreviated as a.e.) if $\mu(\{ x : f(x)\neq 0\})=0$. We denote $\mathcal{M}_{\text{able}}^+=\{ f\in \mathcal{M}_{\text{able}} : f \ge 0\}.$

[Theorem] Any non-negative function $f \in \mathcal{M}_{\text{able}}$ can be approximated by a sequence of simple functions $\{\phi_n\}$, where $\phi_n = \sum_{k=0}^{2^{2n}-1} k 2^{-n} \chi_{E_{n,k}} + 2^n\chi_{F_n}.$ Here, $E_{n,k}=f^{-1}((k2^{-n}, (k+1)2^{-n}])$ and $F_n=f^{-1}((2^n,\infty])$. Moreover, each $\phi_n$ satisfies the inequalities: $$\phi_n \leq \phi_{n+1} \quad \text{and} \quad 0 \leq f(x)-\phi_n(x) \leq 2^{-n}, \quad \forall x\in X\setminus F_n.$$


Definition of the Lebesgue integral

The integral of a measurable simple function $\phi=\sum_{j=1}^n c_j\chi_{E_j}$ is defined as $\int \phi \, d\mu = \sum_{j=1}^n c_j \mu(E_j).$ The integral of $f\in\mathcal{M}_{\text{able}}^+$ is defined by $\int f \, d\mu = \sup \left\{ \int\phi \, d\mu : \phi \leq f \text{ and } \phi\in\mathcal{S}_{\text{imple}} \right\}.$

Based on this definition, we have:
  • Monotone convergence theorem: For a nondecreasing sequence $\{ f_n\}\subset \mathcal{M}_{\text{able}}^+$, we have $\lim_{n}\int f_n \, d\mu = \int \lim_{n} f_n \, d\mu.$
  • For $f\in\mathcal{M}_{\text{able}}^+$, $\int f \, d\mu = 0$ if and only if $f=0$ a.e.
  • Fatou's lemma: For any sequence $f_n\in\mathcal{M}_{\text{able}}^+$, we have $\int \liminf_n f_n \, d\mu \leq \liminf_n \int f_n \, d\mu.$

Banach space $L^1(X,d\mu)$.

A function $f:X\to \mathbb{R}$ is integrable if $f\in\mathcal{M}_{\text{able}}$ and $\int |f| \, d\mu < \infty$. We denote by $L^1(X,d\mu)$ the class of all integrable functions. For $f\in L^1(X,d\mu)$, we define $\int f \, d\mu = \int f^+ \, d\mu - \int f^- \, d\mu.$

Lebesgue dominate convergence theorem

If $\{f_n\} \subset L^1$ with $f_n \to f$ a.e. and there exists $g \in L^1$ such that $|f_n| \leq g$ a.e. for all $n$, then $f \in L^1 \quad \text{and} \quad \int f \, d\mu = \lim_{n}\int f_n \, d\mu.$

Using the Lebesgue Dominated Convergence Theorem, we have $$u(\mathbf{r}) = \nabla^2 \iiint \Phi(\mathbf{r} - \mathbf{r'}) \,  u(\mathbf{r'}) \, d\mathbf{r'}. $$

In the following proof, we utilize the Lebesgue Dominated Convergence Theorem to demonstrate that \(u(\mathbf{r}) = \iiint \Phi(\mathbf{r} - \mathbf{r'}) \, \rho(\mathbf{r'}) \, d\mathbf{r'}\) satisfies the 3D Poisson's equation \(\nabla^2 u = \rho\), thereby confirming it as the solution.

Proof:  Let $\rho \in C^2_0(\mathbb{R}^3)$. A straightforward computation yields
$$-\nabla^2 ( \frac{1}{4\pi|\mathbf{r}-\mathbf{r'}|} ) = 0 \quad \text{if} \quad \mathbf{r} \neq \mathbf{r'}.$$ Hence, for any arbitrary ball $B$ containing $\mathbf{r}$, we have $$\nabla^2 \iiint_{\mathbb{R}^3\setminus B} \frac{-1}{4\pi|\mathbf{r}-\mathbf{r'}|} \rho(\mathbf{r'}) \, d\mathbf{r'} = 0 \quad (\forall B \ni \mathbf{r}).$$
This leads to $$-\nabla^2 \iiint_{\mathbb{R}^3} \frac{1}{4\pi|\mathbf{r}-\mathbf{r'}|} \rho(\mathbf{r'}) \, d\mathbf{r'} = -\nabla^2 \iiint_{B} \frac{1}{4\pi|\mathbf{r}-\mathbf{r'}|} \rho(\mathbf{r'}) \, d\mathbf{r'} \quad (B \ni \mathbf{r}).$$

For $\mathbf{r} \in B$, denote $$ A = -\nabla^2 \iiint_{B} \frac{1}{4\pi|\mathbf{r}-\mathbf{r'}|} \rho(\mathbf{r'}) \, d\mathbf{r'}.$$
Then $A$ can be expressed as $$A = \lim_{\epsilon \to 0+} \iiint_{B} \nabla^2_{\mathbf{r}} ( \frac{-1}{4\pi(|\mathbf{r}-\mathbf{r'}|+\epsilon)} ) \rho(\mathbf{r'}) \, d\mathbf{r'}.$$
Then, $A$ can be decomposed into $$A = \rho(\mathbf{r}) \lim_{\epsilon \to 0+} \iiint_{B} \nabla^2_{\mathbf{r}} ( \frac{-1}{4\pi(|\mathbf{r}-\mathbf{r'}|+\epsilon)} ) \, d\mathbf{r'} + A_2,$$where $$A_2 = \lim_{\epsilon \to 0+} \iiint_{B} \nabla^2_{\mathbf{r}} ( \frac{-1}{4\pi(|\mathbf{r}-\mathbf{r'}|+\epsilon)} ) (\rho(\mathbf{r'}) - \rho(\mathbf{r})) \, d\mathbf{r'}.$$
Given that $$\nabla^2_{\mathbf{r}} \frac{-1}{4\pi(|\mathbf{r}-\mathbf{r'}|+\epsilon)} (\rho(\mathbf{r'}) - \rho(\mathbf{r})) ) \leq C |\mathbf{r}-\mathbf{r'}|^{-2}$$, which is an integrable function in three dimensions, it follows from the Lebesgue Dominated Convergence Theorem, $$A_2 = \iiint_{B} \lim_{\epsilon \to 0+} \nabla^2_{\mathbf{r}} ( \frac{-1}{4\pi(|\mathbf{r}-\mathbf{r'}|+\epsilon)} ) (\rho(\mathbf{r'}) - \rho(\mathbf{r})) \, d\mathbf{r'} = 0.$$
This concludes the proof because
$$\lim_{\epsilon \to 0+} \iiint_{B} \nabla^2_{\mathbf{r}} ( \frac{-1}{4\pi(|\mathbf{r}-\mathbf{r'}|+\epsilon)} ) \, d\mathbf{r'} = -\nabla^2 \iiint_{B} \frac{1}{4\pi|\mathbf{r}-\mathbf{r'}|} \, d\mathbf{r'} = 1 \quad (\forall B \ni \mathbf{r}). $$






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